What is the interpretation of dy/dx in parametric equations and why is it different from the velocity?

Question

So I know normally that dy/dx is equal to the velocity of a particle at a specific point if the original equation indicates the position of that particle. When dealing with parametric equations, I know velocity is equal to <dx/dt, dy/dt>. But it made sense to me that dividing dy/dt over dx/dt, giving dy/dx, would mean the same thing. Of course, it isn't, but I don't understand why it doesn't work and what the actual interpretation of dy/dx is for parametric equations.

Answer 1

These kinds of "fake arithmetic" calculations with derivatives, like $\frac{dy/dt}{dx/dt} = \frac{dy}{dx}$, are fraught with danger. Don't let your facility at cancelling fractions fool you into thinking that the result of that cancellation has the physical meaning that you might think it has, not without some very careful thought.

But it is still a fair question to ask what physical significance the ratio $\frac{dy/dt}{dx/dt}$ might have.

When dealing with a moving particle whose position vector at each moment of time $t$ is given by $\langle x(t), y(t) \rangle$, its velocity vector is indeed $\langle x'(t), y'(t) \rangle$, which can also be written as you expressed it, namely $\langle dx/dt, dy/dt \rangle$.

Keeping in mind that $\langle dx/dt, dy/dt \rangle$ is the velocity vector, we are certainly allowed to consider its two component functions $x'(t)=dx/dt$ and $y'(t)=dy/dt$ separately, and to study that vector by doing some arithmetic operations on those components. I can think of two important quantities obtained in this fashion:

• The speed of the particle at time $t$ is the magnitude of the velocity vector, namely $$\sqrt{(x'(t))^2 + (y'(t))^2} $$
• The slope of velocity vector at time $t$ is the ratio of its coordinates $$\frac{y'(t)}{x'(t)} = \frac{dy/dt}{dx/dt} $$ which is the quantity in your question.

But it is never correct to refer to that ratio $\frac{dy/dt}{dx/dt}$ as the velocity of the particle, despite the "fake arithmetic" calculation $\frac{dy/dt}{dx/dt} = \frac{dy}{dx}$:

• The velocity of a particle moving in a plane with position function $\langle x(t),y(t) \rangle$ is a vector with 2 components, namely $\langle x'(t),y'(t) \rangle$.
• The velocity of a particle moving along a line with position function $f(t)$ is a single number $f'(t)$. You can also think of this as a vector with 1 component, namely $\langle f'(t) \rangle$.

Nonetheless, that "fake arithmetic" calculation is a clue to something interesting, namely:

The Implicit Function Theorem: Consider a particle in the plane with position function $\langle x(t),y(t) \rangle$ at time $t$. Given $t_0$, if $x'(t_0) \ne 0$ then there exists a function $y=f(x)$ and a real number $r > 0$ such that $$y(t)=f(x(t)) \quad\text{if $t_0-r < t < t_0+r$} $$ It follows (from the chain rule) that $$y'(t) = f'(x(t)) \, x'(t) $$ and therefore $$f'(x(t)) = \frac{y'(t)}{x'(t)} $$

And there, on the right hand side, you can see your ratio $\frac{dy/dt}{dx/dt} = \frac{y'(t)}{x'(t)}$. So from this we can derive an interesting physical interpretation of the ratio $\frac{dy/dt}{dx/dt}$. Namely, under the conclusions of the Implicit Function Theorem, the track of the moving particle is given by the graph of a function $y=f(x)$, and your ratio is equal to the slope of the graph of that function, which is simply the derivative of that function with respect to $x$.

Answer 2

Both of the following describe a curve in 2D ($\mathbb{R}^2$):

$$y = f(x)$$

and

$$(x,y) = (g(t), h(t))$$

So the question is "what is velocity?" Well it all depends on how you interpret your variables. Normally, velocity is thought of as the change in position with respect to time. Let's call time $t$. Notice how time doesn't appear anywhere in $y = f(x)$. Let's fix that by rewriting it this way:

$$ (x,y) = (t, f(t)) $$

This gives us the same curve, but now it's expressed parametrically in terms of time. So what's the velocity? Well:

$$ \frac{dx}{dt} = 1 $$

and $$ \quad \frac{dy}{dt} = \frac{df(t)}{dt} = \frac{df(x)}{dx} = \frac{dy}{dx} $$

So when you say that $dy/dx$ is a velocity, what you really mean is that it is the $y$-component of the velocity ($x$ changes too!) with the assumption that $x$ has a velocity of 1.

That is how these two ideas are related; people just never make it clear what they mean by velocity in the case $y=f(x)$, which can cause a lot of confusion.

Answer 3

At the point $(x,y),\ \ \frac{dy/dt}{dx/dt} = \frac{\text{vertical component of velocity}}{\text{horizontal component of velocity}} = \tan\theta,\ $ where $\theta$ is the angle that the instantaneous velocity vector at the point $(x,y)$ makes anticlockwise from the positive $x-$axis.

So, $\ \arctan\left(\frac{dy/dt}{dx/dt}\right)$ is the direction of the velocity vector.

In general I don't think $\frac{dy/dt}{dx/dt}$ is related to the magnitude of the velocity vector.